4 Analytic Trigonometry 69 Chapter 4 Analytic Trigonometry 4.1 Inverse Trigonometric Functions The trigonometric functions act as an operator on the variable (angle) x, resulting in an output value y. Suppose this process is reversed: given a y-value, is it possible to work backward to determine the angle x that produced y? In simplest terms, we wish to solve equations such as . , 1 sec , tan , sin 5 2 2 1 etc x x x In some cases the value x can be determined “by inspection”. For example, the simple algebraic equation 2 1 sin x implies that 6 x is one possible solution. However, this method does not work this easy in general: for example, by inspection what is the solution to 5 2 tan x ? Review of Inverse Graphs and Inverse Functions A function ) ( x f y generates a set of points )} ( | ) , {( x f y y x that are plotted on a Cartesian (x-y) coordinate axis system. The result is the graph of the function. The inverse graph of a function ) ( x f y is a graph of the set of points )} ( | ) , {( x f y x y . In simplest terms, the inverse graph of ) ( x f y is a new graph in which each ordered pair ) , ( y x of ) ( x f y has its coordinates reversed to ) , ( x y . The points ) , ( x y and ) , ( y x are symmetrical across the line x y ; this allows a simple way to sketch the inverse graph of a function ) ( x f y . However, in most cases the inverse graph will not be a function since it fails the vertical line test. To ensure that the inverse graph of a function ) ( x f y is itself a function, the given function must be one-to-one. A function is one-to-one if ) ( ) ( b f a f implies that b a . Visually, a function that is one-to-one passes the horizontal line test. To summarize, a function that passes the horizontal line test is said to be one-to-one, and if this condition is met, its inverse graph will be a function as well. If this is true, then the inverse function is denoted as ) ( 1 x f y . The Inverse Sine Function (Arcsine) Consider the function x x f y sin ) ( . Its graph is given in the next page:
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4 Analytic Trigonometry
69
Chapter 4 Analytic Trigonometry
4.1 Inverse Trigonometric Functions
The trigonometric functions act as an operator on the variable (angle) x, resulting in an
output value y. Suppose this process is reversed: given a y-value, is it possible to work
backward to determine the angle x that produced y?
In simplest terms, we wish to solve equations such as
.,1sec,tan,sin52
21 etcxxx
In some cases the value x can be determined “by inspection”. For example, the simple
algebraic equation 21sin x implies that
6x is one possible solution. However, this
method does not work this easy in general: for example, by inspection what is the
solution to 52tan x ?
Review of Inverse Graphs and Inverse Functions
A function )(xfy generates a set of points )}(|),{( xfyyx that are plotted on a
Cartesian (x-y) coordinate axis system. The result is the graph of the function.
The inverse graph of a function )(xfy is a graph of the set of points
)}(|),{( xfyxy . In simplest terms, the inverse graph of )(xfy is a new graph in
which each ordered pair ),( yx of )(xfy has its coordinates reversed to ),( xy . The
points ),( xy and ),( yx are symmetrical across the line xy ; this allows a simple way
to sketch the inverse graph of a function )(xfy . However, in most cases the inverse
graph will not be a function since it fails the vertical line test.
To ensure that the inverse graph of a function )(xfy is itself a function, the given
function must be one-to-one. A function is one-to-one if )()( bfaf implies that
ba . Visually, a function that is one-to-one passes the horizontal line test.
To summarize, a function that passes the horizontal line test is said to be one-to-one, and
if this condition is met, its inverse graph will be a function as well. If this is true, then the
inverse function is denoted as )(1 xfy .
The Inverse Sine Function (Arcsine)
Consider the function xxfy sin)( . Its graph is given in the next page:
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The sine function is not one-to-one since it does not pass the horizontal line test.
However, if the domain of xxf sin)( is restricted to the interval 22
x , then it is
one-to-one and its inverse graph is therefore a function:
Therefore, the inverse sine function (also called the arcsine function, written arcsin(x)),
is defined to be the inverse graph of the function xxf sin)( on the closed interval
22x . The domain of the inverse sine function is 11 x and the range is
22)(xf . The inverse sine function returns values in the 1
st quadrant (if 0x ) or
the 4th
quadrant (if 0x ).
Example 1: Determine values for (a) 21arcsin , (b)
2
3arcsin , (c) 2
2arcsin ,
(d) )1arcsin( and (e) )arcsin( .
Solutions: The results for (a), (b) and (c) are based on the normal measures of the sine
function in the first and fourth quadrants. Thus, 62
1arcsin , 32
3arcsin and
42
2arcsin . For (d), 2
)1arcsin( and for (e), the solution is not defined since is
outside the domain.
Example 2: Using a calculator set to degree mode, what is )4.0arcsin( ?
Solution: The solution to two decimal places is the angle 23.58 degrees.
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Example 3: Using a calculator in radian mode, what is )arcsin(61 ?
Solution: To two decimal places, the result is 0.17 radian.
The Inverse Cosine Function (Arccosine)
The cosine function )cos()( xxf is not one-to-one over its usual domain of the Real
numbers. However, when restricted to the closed interval x0 , the cosine function
is one-to-one, and hence its inverse graph is also a function, which will be defined at the
inverse cosine function (also written )arccos(x ).
The inverse cosine graph )arccos()( xxf is defined on the domain 11 x with a
range of )(0 xf . It returns values in the 1st quadrant (if 0x ) or the 2
nd quadrant
(if 0x ).
Example 4: Determine values for (a) 21arccos , (b)
2
3arccos , (c) 2
2arccos , (d)
)1arccos( and (e) )arccos( .
Solutions: The results for (a), (b) and (c) are based on the normal measures of the cosine
function in the first and second quadrants. Thus, 32
1arccos , 62
3arccos and
43
2
2arccos . For (d), 0)1arccos( and for (e), the solution is not defined since
is outside the domain.
Example 5: What radian measure solves )215.0arccos( ?
Solution: The answer is 1.354 radians.
Example 6: If 32)sin(x , find )cos(x .
Solution: This is identical to the question ))(cos(arcsin32 . On a right triangle, the
opposite measure of one of the non-right angles is 2 and the hypotenuse is 3. With these
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facts, the Pythagorean Theorem allows for the adjacent leg to be solved: 523 22.
Therefore, the cosine of this angle is 3
5 .
The Inverse Tangent Function (Arctangent)
Like the sine and cosine function, the tangent function )tan()( xxf is not one-to-one
unless restricted to a smaller domain, which is usually chosen to be 22
x . Thus
restricted, the inverse tangent function (written )arctan(x ) is defined as the inverse
graph of the tangent function.
The domain of the arctangent function is x and its range is 22
)(xf . It
returns values in the 1st quadrant (if 0x ) or the 4
th quadrant (if 0x ). Notably, it has
two horizontal asymptotes at 2
y as x , and 2
y as x .
Example 7: Determine the values to (a) )3arctan( , (b) )1arctan( and (c) )arctan(2
.
Solutions: For (a), 3
)3arctan( , for (b), 4
)1arctan( and for (c), the result is
undefined, although it trends to positive infinity as 2
x .
Example 8: Evaluate ))(sin(arctan41 .
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Solution: The arctan of 41 is an angle whose opposite leg is 1 and adjacent leg is 4.
Therefore, the hypotenuse is 1741 22. Thus, the sine of this angle is the opposite
divided by the hypotenuse: 17
17
17
141 ))(sin(arctan .
Important Results
1 1 1 1 1 1sin ( ) sin ( ), tan ( ) tan ( ), cos ( ) cos ( )t t t t t t
Example 9: Evaluate the following expressions
a) 1sin (1) b) 1sin ( 1) c) 1sin ( 1/ 2) d) 1cos (1) e) 1cos ( 1)
f) 1 2cos
2 g)
1tan (1) h) 1tan ( 3) i) 1 3tan
3
Solution:
a) 1 1sin (1) sin (sin( / 2)) / 2
b) 1 1sin ( 1) sin (1) / 2
c) 1 1sin ( 1/ 2) sin (1/ 2) sin(sin( / 6)) / 6
d) 1 1cos (1) cos (cos0) 0
e) 1 1cos ( 1) cos (1) 0
f) 1 12 2cos cos / 4 3 / 4
2 2
g) 1 1tan (1) tan (tan( / 4)) / 4
h) 1 1 1tan ( 3) tan ( 3) tan (tan( /3)) / 3
i) 1 1 13 3
tan tan tan tan( / 6) / 63 3
Common Errors to Avoid
Students sometimes make the erroneous assumption that the “arcsin” cancels the “sin”,
for example, xx))(arcsin(sin , which is not necessarily true! Pay special attention to
the domain appropriate to the particular function being evaluated. Consider these
following examples:
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Example 10: Is it true that 1010
))(arcsin(sin ?
Solution: Yes, since 2102
.
Example 11: Is it true that 4
54
5 ))(arcsin(sin ?
Solution: No. The argument 4
5 is outside the domain of arcsine. The arcsine function
returns the smallest such angle corresponding to the input. Since 4
5 lies in the 3rd
quadrant, the arcsine function will return an angle in the 4th
quadrant equivalent to 4
5 .
The correct result is 44
5 ))(arcsin(sin .
Example 12a: Is it true that 4
34
3 ))(arccos(cos ?
Solution: Yes. The argument 4
3 is within the domain of the arccosine function
430 .
Example 12b: Is it true that 4
34
3 ))(cos(arccos ?
Solution: No. The value 4
3 is outside the domain 11 x of the arccosine function.
This expression is undefined.
Generalized Expressions
The argument can be left as an independent variable and expressions combined to derive
relationships between the sine, cosine and tangent operations with the arcsine, arccosine
and arctangent inverse operations.
Example 13: Simplify the expression ))(cos(arcsin x .
Solution: The )arcsin(x suggests a result y such that xy)sin( . Sketching a right
triangle with x at the opposite leg relative to angle y, and 1 at the adjacent leg to y, the
Pythagorean Theorem gives the hypotenuse as having length 12x . Therefore, the
cosine of this angle (y) is the adjacent leg divided by the hypotenuse:
1
1))(cos(arcsin)cos(
2xxy .
Example 14: Simplify the expression ))(arccos(sin3
Solution: Since 63
cossin , rewrite ))(arccos(sin3
as ))(arccos(cos6
. Since 6
is on
the domain of the arccosine function, the expression reduces to 66
))(arccos(cos .
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Example 15: Solve the equation 01sinsin2 2 xx in the interval 20 x .
Solution: Factor the equation:
0)1)(sin1sin2(
01sinsin2 2
xx
xx
Set each factor equal to zero and solve for x:
621
21 )arcsin(sin01sin2 xxxx
2)1arcsin(1sin01sin xxxx
Note that the arcsine function returns values in quadrants 1 and 4 only. Furthermore,
other solutions may need to be inferred from the ones provided by the arcsine operation:
Since 6
x is a solution, there also exists a solution in quadrant 2 by symmetry.
Therefore, 6
5x is also a solution to the equation.
The solution 2
x provided by the arcsine operation is correct but outside the desired
bounds 20 x . This is easily remedied by selecting 2
3x as the equivalent
solution.
Therefore, the solution set to this equation is },,{2
36
56
.
Example 16: Solve the equation 03tan4tan2 xx in the interval 22
x .
Solution: Factor the equation:
0)1)(tan3(tan
03tan4tan 2
xx
xx
Each factor is set equal to zero and solved for the variable:
radiansxxxx ...249.1)3arctan(3tan03tan
radiansxxxx4
)1arctan(1tan01tan
The solution set (in radians) is },249.1{4
.
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4.2 Trigonometric Identities
An identity is an equation that is always true for all values in its domain. For example,
the equation ))((22 bababa is an identity since it is true for all a and b. In a
similar manner, many trigonometric expressions can be expressed in different forms, thus
forming trigonometric identities.
Reciprocal Identities
From the definitions of the six trigonometric functions, the tangent, cotangent, secant and
cosecant functions can all be expressed as identities involving the sine and/or cosine
functions:
xx
xx
x
xx
x
xx
sin
1csc
cos
1sec
sin
coscot
cos
sintan
Simple algebraic manipulations can generate more identities. As an example, since
x
xx
cos
sintan , it is possible to generate a new identity by multiplying the cos x and
deriving xxx costansin . However, restrictions placed on the domain x are
maintained throughout any further manipulations. Thus, the expression
xxx costansin is true as long as Nnxn
,2
)12(, since these restrictions are in
place because of the presence of the tangent function.
The Pythagorean Identities
The sine and cosine functions are defined on the unit circle and are related by the
Pythagorean identity:
1cossin 22 xx
With simple algebra, new corollary identities can be formed:
xx
xx
22
22
sin1cos
cos1sin
These identities are true for all x .
Furthermore, the Pythagorean Identity can be divided through by x2sin or x2cos to
generate more identities:
Divide by x2sin :
xxxx
x
x
xxx 22
22
2
2
222 csccot1
sin
1
sin
cos
sin
sin1cossin
Divide by x2cos :
xxxx
x
x
xxx 22
22
2
2
222 sec1tan
cos
1
cos
cos
cos
sin1cossin
These identities are true for all x for which the expressions are defined.
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General Identities and Demonstrations
The purpose of employing identities is to reduce a large trigonometric expression into
something smaller and easier to manipulate. The next few examples illustrate the method
to demonstrate the truthfulness of each identity. Until the proof is demonstrated, the
(potential) identity is called a theorem.
The general method is to use whatever identity is appropriate at any given step – this
often takes practice to “recognize” the right identity. Furthermore, there may be more
than one way to demonstrate the truthfulness of the identity. The only requirement that
must not be violated is that each side of the identity must be handled separately. Any
algebraic maneuver that assumes the equality is true cannot be used since the truthfulness
of the equation has yet to be shown! In other words, you cannot use what you are trying
to prove. In most cases, this removes the method of cross-multiplication from
consideration. Otherwise, normal algebraic techniques are used as needed. Always work
from the most complicated form first.
Example 1: Prove that xx
x
x
x
x
x
cossin
cos1
cos
sin
sin
cos1
Solution: The left side of the theorem shows two rational expressions being summed.
Rewrite the left side as a sum by using a common denominator. The common
denominator of xsin and xcos is their product, xxcossin . The numerators must be
adjusted accordingly:
xx
xxx
x
x
x
x
x
x
x
x
cossin
sincos)cos1(
sin
sin
cos
sin
cos
cos
sin
cos1 2
The expression on the right side can be simplified by distributing the xcos :
xx
xxx
xx
xxx
cossin
sincoscos
cossin
sincos)cos1( 222
.
Since 1cossin 22 xx , the numerator reduces to
xx
x
xx
xxx
cossin
1cos
cossin
sincoscos 22
Thus, we have shown by transitivity the following:
xx
xxx
x
x
x
x
x
x
x
x
x
x
x
x
cossin
sincos)cos1(
sin
sin
cos
sin
cos
cos
sin
cos1
cos
sin
sin
cos1 2
xx
x
xx
xxx
cossin
1cos
cossin
sincoscos 22
Therefore, the left side of the theorem is equal to the right side. The theorem is proven; it
is an identity.
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Example 2: Prove that x
x
xx
xx
sec
sin
cscsec
cossin
Solution: The left side of the theorem contains “more” expressions with which to
manipulate. Convert all functions into equivalent forms in terms of sine and cosine:
xx
xx
xx
xx
sin1
cos1
cossin
cscsec
cossin
The denominators of the two small expressions in the denominator of the main
expression can be combined by summing over a common denominator:
xxxx
xxx
xxx
xx
xxxxxx
cossincossin
cossincos
cossinsin
sin1
cos1
cossincossincossin
Reciprocating the combined expression in the main denominator, the “ xx cossin ”
expressions cancel:
xxxx
xxxx
xx
xxxx
cossincossin
cossin)cos(sin
cossin
cossincossin
Since x
xsec
1cos , the expression xxcossin now becomes
x
x
xxxx
sec
sin
sec
1sincossin
Thus, the theorem is proven.
Notice that we never once manipulated the expression xx
secsin directly in this proof. We
were able to show the relationship entirely by manipulating the expressions from the left
side of the original theorem.
Shift and Reflection Identities
The sine and cosine functions are identical in shape and differ only by a horizontal shift.
Specifically, the sine function is the cosine function shifted 2
units to the right.
Conversely, the cosine function is the sine function shifted 2
units to the left. Therefore,
two useful identities can be stated:
)sin(cos
)cos(sin
2
2
xx
xx
Furthermore, the sine function shifted units to the left or right results in an inverted
sine function. The same is true for the cosine function. These two identities are
summarized below:
xx
xx
cos)cos(
sin)sin(
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Shifting the sine and cosine functions 2
units in opposite directions result in inverted
cosine and sine functions:
xx sin)cos(2
xx cos)sin(2
Of course, the sine and cosine functions have a period of 2 . Therefore, two more
identities can be stated:
xx
xx
cos)2cos(
sin)2sin(
The tangent function has a period of so its shift identity is
xx tan)tan(
Shift identities for the secant, cosecant and cotangent functions can be derived by
converting into sine and cosine functions.
A vertical reflection of any function can be determined by replacing the argument x with
x . The result is a graph that is reflected across the y-axis. The cosine function is
already symmetrical to the y-axis so it will remain unchanged. The sine and tangent
functions, both symmetrical to the origin, will invert across the x-axis. Thus, the
following reflection identities can be stated:
xx
xx
xx
tan)tan(
sin)sin(
cos)cos(
For now, the “proofs” of these identities are done visually. In the next section, a method
will be developed to analytically prove all such shift and reflection identities.
Example 3: Prove that xx cos)sin(2
.
Solution: Factor the negative within the argument of the sine function:
))(sin()sin(22
xx
The leading negative within the argument can be “moved” to in front of the sine function:
)sin())(sin(22
xx
Since xx cos)sin(2
, then xxx cos)cos()sin(2
.
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4.3 Sum and Difference Formulas
The sum and difference forms of the sine and cosine functions are the following four
forms:
)sin(),cos(),sin( yxyxyx and )cos( yx .
The trigonometric function can be viewed as operators on the variable(s) x and y, but they
are not linear operators. The function/operators cannot be distributed across addition and
subtraction. Hence, statements like yxyx sinsin)sin( are false. Unfortunately
these are common mistakes students make when considering the sum and difference
forms of the trigonometric functions.
In this section we will present a geometric proof for all these four forms.
We prove that
sin( ) sin cos cos sinx y x y x y
Suppose that OX be the initial side and the
terminal side OY makes an angle x with initial
side and the terminal side OZ makes angles y
with the side OY. Take a point P on the side
OZ and make perpendicular lines PQ and PS
on OX and OY respectively. Further we make
perpendicular line ST and SR according to the
adjacent diagram. Now observe that the angle
TPS is equal to the angle x. The angle POR is
equal to x + y.
Z
P
T x Y
S
y
x
O Q R X
sin( )
cos sin sin cos
PQ PT TQx y
OP OP
PT TQ PT SR
OP OP OP OP
PT PS SR OS
PS OP OS OP
x y x y
And
sin( ) sin cos( ) cos sin( )
sin cos cos sin , sin( ) sin , cos( ) cos
x y x y x y
x y x y x x x x
Now we have to prove the result for cosine:
Remember that
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cos( ) sin ( ) sin sin cos( ) cos sin( )2 2 2 2
cos cos sin sin , sin( ) sin , sin cos , cos( ) cos2
x y x y x y x y x y
x y x y x x x x x x
Also cos( ) cos cos( ) sin sin( ) cos cos sin sinx y x y x y x y x y
We now present the algebraic proof.
The sum and difference forms can be derived from an analysis of the angles drawn on a
unit circle, and the familiar distance formula. We consider the case )cos( yx first, from
which the other three forms can easily be derived as corollaries.
Proof of
cos(x – y) = cos x cos y + sin x sin y
Consider the unit circle in the first quadrant (without loss of generality). Let ray xr be
drawn with angle x, and yr be drawn with angle y, and let yx . Therefore, ray xr
intersects the circle at point )sin,(cos xx and ray yr intersects the circle at point
)sin,(cos yy .
Rigidly rotate the circle clockwise through an angle of y, so that the original ray yr sits
along the positive x-axis while original ray xr now has an angle of elevation yx and
therefore intersects the unit circle at the point ))sin(),(cos( yxyx .
The core of the proof is to show that the distance between points )sin,(cos xx and
)sin,(cos yy is the same as the distance from point ))sin(),(cos( yxyx to the point
)0,1( . See the following diagram:
Figure. Diagram showing rays xr and yr (left), and the rotation through an angle y (right).
The distance between )sin,(cos xx and )sin,(cos yy is found by the distance formula:
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yxyx
yyxxyyxx
yxyxyxyxD
sinsin2coscos22
sinsinsin2sincoscoscos2cos
)sin(sin)cos(cos))sin,(sin),cos,((cos2222
22
The Pythagorean identity 1cossin 22 xx was used twice in the second-to-third step.
The distance between ))sin(),(cos( yxyx and (1,0) is
)cos(22
)(sin1)cos(2)(cos
)0)(sin()1)(cos())0,1()),sin(),((cos(22
22
yx
yxyxyx
yxyxyxyxD
The Pythagorean identity 1)(cos)(sin 22 yxyx was used in the second-to-third
step.
Since the two distances are equal, relate them by equality:
yxyxyx sinsin2coscos22)cos(22
Squaring both sides removes the radicals. The constants cancel, and the above equation
algebraically reduces to:
yxyxyx sinsincoscos)cos( (A)
Therefore, the difference formula for the cosine function is proved.
Proofs of the Remaining Forms
The other three forms can be derived using the difference form of the cosine function (A)
above. For example, using the shift identities xx sin)cos(2
and xx cos)sin(2
,
we can make the substitution 2
x for x into the above form (A) to get
yxyxyx sin)sin(cos)cos()cos(222
(B)
The left side of (B) can be re-arranged as )cos()cos(22
yxyx , which equals
)sin( yx . On the right side of (B), we use the substitutions xx sin)cos(2
and
xx cos)sin(2
. Therefore, equation (B) can be written as
yxyxyx sincoscossin)sin(
This proves the difference formula for the sine function.
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The sum forms can be proven using the symmetry identities xx sin)sin( and
xx cos)cos( . Therefore, the term y is substituted with y in equation (A) to get
yxyxyx
yxyxyx
sinsincoscos)cos(
)sin(sin)cos(cos))(cos(
This proves the sum form for the cosine function.
The same substitution is made in equation (B):
yxyxyx
yxyxyx
sincoscossin)sin(
)sin(cos)cos(sin))(sin(
This proves the sum form for the sine function.
The Sum and Difference Identities for the Sine and Cosine Functions
yxyxyx
yxyxyx
yxyxyx
yxyxyx
sinsincoscos)cos(
sincoscossin)sin(
sinsincoscos)cos(
sincoscossin)sin(
The following examples illustrate some of the ways these formulas can be used:
Example 1: Calculate )sin(12
exactly.
Solution: The angle 12
is the difference of angles 3
and 4
: 4312
. Therefore,
4
26
2
2
21
2
2
2
3
43434312))(())(()sin()cos()cos()sin()sin()sin(
Example 2: Calculate )105cos( exactly.
Solution: 105 degrees can be written as the sum of 60 and 45 degrees. Therefore,